Assume that \(X\cong P\Omega _{n}^{-}(q)\)  with \(n \geqslant 6\) . Actually, \(P\Omega _{6}^{-}(q)\)  cannot occur by Lemma REF , since \(P\Omega _{6}^{-}(q)\cong PSU_{4}(q)\) . Thus \(n \geqslant 8\) . The group \(X_{x}\)  lies in a large maximal
geometric subgroup \(Y\)  of \(X\)  by Theorem REF . Then either \(Y\in \mathcal {C}_{1}(X)\) , or \(Y\)  is a \(\mathcal {C}_{3}\) -subgroup of \(X\)  either of type \(O_{n/2}^{-}(q^{2})\)  or \(GU_{n/2}(q)\)  according as \(n/2\)  is even or
odd, respectively, by [1]}, Proposition 4.23, since \(\left( \Phi _{nf}^{\ast }(p),\left|Y\right|\right) >1\) . Actually, the case \(Y\in \mathcal {C}_{1}(X)\)  is ruled out by Theorem 3.5 of [2]}. Thus,
either \(Y\cong P\Omega _{n/2}^{-}(q^{2}).Z_{2}\)  or \(Y\cong \frac{q+1}{\left(q+1,4\right) }.PSU_{n/2}(q).\left[ \left( q+1,n/2\right) \right] \)  by [3]}, Propositions 4.3.16.(II) and 4.3.18.(II). Moreover, \(X_{x}=Y\)  by Lemma REF (1) and by [3]}, Propositions 4.3.16.(I) and 4.3.18.(I).
Assume that \(X_{x}\cong P\Omega _{n/2}^{-}(q^{2}).Z_{2}\) . Then
\(k^{2}=\frac{1}{2}q^{n^{2}/8}\prod _{i=0}^{n/4-1}(q^{4i+2}-1)\text{.}\) 
Then \(n^{2}/8\)  is even, as \(n/2\)  is even. If \(q\)  is even, then the
highest power of 2 dividing \(k^{2}\)  is \(2^{f\frac{n^{2}}{8}-1}\) , but this is a
contradiction as \(fn^{2}/8-1\)  is odd. Therefore, \(q\)  is odd and hence (REF ) becomes
\(k^{2}=q^{n^{2}/8}\frac{(q^{2}-1)^{n/4}}{2}\prod _{i=1}^{n/4-1}\left( \frac{q^{4i+2}-1}{q^{2}-1}\right) \text{.} \) 
Moreover, the actions
of \(X\)  on a suitable \(X\) -orbit of spreads of \(PG_{n-1}(q)\)  and on the point set of \(\mathcal {D}\) 
are equivalent (see above). Thus we may identify the two sets.
Suppose that \(n/4\)  is even and let \(2^{j}\)  be the highest power of 2
dividing \(q^{2}-1\) . Then \(2^{jn/4-1}\)  must be a square, as \(q^{n^{2}/8}\prod _{i=1}^{n/4-1}\left( \frac{q^{4i+2}-1}{q^{2}-1}\right) \)  is odd, whereas it is not, as \(jn/2-1\)  is odd, and so we reach a contradiction. Thus \(n/4\)  is odd and hence \(n\geqslant 12\) , as \(n \geqslant 8\) .
Since there is \(\alpha \in X-X_{x}\)  such that \(\Omega _{n/2-1}(q^{2})\leqslant X_{x,x^{\alpha }}\)  by Lemma REF , it follows that \(\Omega _{n/2-1}(q^{2})\leqslant G_{x,x^{\alpha }}\) . Thus \(k+1 \mid 8q^{n/2-2}(q^{n/2}+1)f\) , as \(k+1\)  divides \(\left[G_{x}:G_{x,x^{\alpha }}\right] \)  by Lemma REF (3), and hence \(k+1\mid 8(q^{n/2}+1)f\) , as \(p\)  divides \(k\) . So,
\(q^{n^{2}/8}(q^{2}-1)^{n/4}<k<64(q^{n/2}+1)^{2}f^{2}<16(q^{n/2}+1)^{2}q^{2}\text{,}\) 
which has no solutions for \(n\geqslant 12\) .
Assume that \(X_{x}\cong \frac{q+1}{\left( q+1,4\right) }.PSU_{n/2}(q).\left[\left( q+1,n/2\right) \right] \) . Let \(\mathcal {Q}\)  be the elliptic quadric
of \(PG_{n-1}(q)\)  preserved by \(X\) . Then the actions of \(X\)  on the sets of the spreads of lines
of \(\mathcal {Q}\)  and on the point set of \(\mathcal {D}\)  are equivalent. Therefore, we may identify the points of \(\mathcal {D}\)  with the spreads of lines of the elliptic quadric \(\mathcal {Q}\) . Hence
\(k^{2}=q^{n(n-2)/8}\prod _{i=2}^{n/2-1}\left( q^{i}+\left( -1\right)^{i}\right) \text{,}\) 
where \(n\geqslant 10\) , as \(n/2\)  is odd and \(n\geqslant 8\) .
There is \(\beta \in X-X_{x}\)  such that \([X_{x}:X_{x,x^{\beta }}]\)  divides \(2(q+1,n/2)(q+1,n/2-2)q^{n-3}(q^{n/2}+1)(q^{n/2-1}-1)\)  by Lemma REF , since \(n/2\)  is odd. Then
\(k+1 \mid f(q+1,n/2)(q+1,n/2-2)(q^{n/2}+1)(q^{n/2-1}-1)\) 
by Lemma REF (3), since \(k\)  is even and \(p\)  divides \(k\) . Now, comparing (REF ) and (REF ), and using [1]}, Lemma 4.1(i), we obtain
\(q^{n(n-2)/4-1 }<k^{2}<q^{2}(q+1)^{4}(q^{n/2}+1)^{2}(q^{n/2-1}-1)^{2}<64q^{2n+4}\) 
and hence \(n=10\) , as \(n/2\)  is odd. Then (REF ) becomes
\( k^{2}=q^{10}(q^{2}+1)(q^{3}-1)(q^{4}+1)\) 
which, together with (REF ), yields \(k+1 \mid f(q+1,15)(q^{5}+1)(q+1)\) . Then \(k^{2}<225q^{2}(q+1)^{2}(q^{5}+1)^{2}\)  which compared to REF  yields \(q=2,3\) . However, these values of \(q\)  do not fulfill (REF ), and hence this case is ruled out. This completes the
proof.
